The area (in sq unit) of the region bounded b | vedclass

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(B) The given curve is $2x = y^2 - 1$,which can be rewritten as $x = \frac{y^2 - 1}{2}$.The curve $x = 0$ represents the $y$-axis.The intersection poi

Vedclass · Accepted Answer

$\frac{2}{3}$

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(B) The given curve is $2x = y^2 - 1$,which can be rewritten as $x = \frac{y^2 - 1}{2}$.The curve $x = 0$ represents the $y$-axis.The intersection points of the curve $x = \frac{y^2 - 1}{2}$ and $x = 0$ are found by setting $\frac{y^2 - 1}{2} = 0$,which gives $y^2 = 1$,so $y = \pm 1$.The area bounded by the curve and the $y$-axis is given by the integral of $|x|$ with respect to $y$ from $y = -1$ to $y = 1$.$	ext{Area} = \int_{-1}^{1} |x| \, dy = \int_{-1}^{1} \left| \frac{y^2 - 1}{2} ight| \, dy$.Since the region is symmetric about the $x$-axis,the area is $2 \int_{0}^{1} \left| \frac{y^2 - 1}{2} ight| \, dy$.In the interval $[0, 1]$,$y^2 - 1 \leq 0$,so $|\frac{y^2 - 1}{2}| = -\frac{y^2 - 1}{2} = \frac{1 - y^2}{2}$.$	ext{Area} = 2 \int_{0}^{1} \frac{1 - y^2}{2} \, dy = \int_{0}^{1} (1 - y^2) \, dy$.$= [y - \frac{y^3}{3}]_{0}^{1} = (1 - \frac{1}{3}) - (0 - 0) = \frac{2}{3} 	ext{ sq unit}$.

The area (in sq unit) of the region bounded by the curves $2x = y^2 - 1$ and $x = 0$ is

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